cg,ca,ci,cc | on ElectriciansForums
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Discuss cg,ca,ci,cc in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

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lofty84

a single phase load of a 13a is supplied via a 15 a bs 1361 fuse using single core 70 degree c theromplastic copper cables installed to method 4 (ref method B) the rating factor for grouping is 0.7 and for ambient temperature 0.87 and overload protection is to be provided. the minimum acceptable size cable would be

a)1mm sq
b)1.5mm sq
c)2.5mm sq
d0 4 mm sq

ive used the brb and its told me to go to app 4

I know In is going to be 15


so 15 / ca ci cc

ca 0.87
ci no idea
cc 0.9

this gives me this for an answer but its obviously wrong


15/0.87x0.9 = 19.16


can anyone show me where im going wrong please
 
There appears to be a reference method of 4 or A which will derate the cable from you maximum clipped direct value. (from your tutor's example :) )
There will not be Cc (see 433.1.1 ,2 ,3)
You have missed out Cg = 0.7. Since they are all running concurrently I would have thought you include the lot in one formula.

Try
In/(Cg*Ca) then find a suitable sized cable from the correct table and column in appendix 4.
 
As your tutor has not mentioned multiple circuits I would not bother with Cg.

I don't think you are wrong with your figure of 19.16, but you seem to be forgetting that the total is an Ampere total not a CSA one. So all you need to do now is to look up which of the cables listed is capable of carrying 19.16A
 
cheers guys I did try it ur way first ack but it didnt sound right to me I then tried it this way and thought I had the answer but I forgot it was in A i would need to find a cable capable of carring it. the thing is where the bloody hell do I look in the brb for it, im looking from pg 274 onwards but cant find what im after. this is going to take more then 2 mins if i get one like it in the test later
 
Doh, sorry the tutor did mention Cg. Overlooked that.

Lets try again...

15A / Ca of 0.87

17.24A / Cg of 0.70

24.63A

Don't think Cc applies in this case as the cable is not buried in the ground (but I'm probably wrong).

Now check Table 4D2A.
 
Its reasonably easy when you get the hang of it.

Know the type of cable - T&E which could be found in two places, 4D2A or 4D5.

4D2A used to be the table for T&E now flat cables in domestic environments are dealt with in 4D5 and has a reference back to 4D2A.

So column 7 4D5 = 20A min for 2.5mm2, 26A for 4.0mm2 , 32A for 6.0mm2
and Column 2 4D2A = 18.5A for 2.5mm2, 25A for 4.0mm2, 32A for 6.0mm2

In/(Ca*Cg) or 15/0.87 *0.7 = 24.6A

So 4.0mm2 will give overload and short circuit protection .

When do I get my certificate? - BTW Jamesbrownlive doesn't get one since he missed out the grouping factor :)
 
Last edited:
An industrial Board supples

- 4 TP motors
- 2 x TP&N motors
- 4 x single phase ring circuits
- 2 x single phase lighting circuits

how many live conductors are there

a)36
b)38
c)40
d)52

anyone know how to work that out ...
 
First of all where's my cert?

Ask yourself, what does it mean by 'live conductors' ?

Then, usually from experience you will know how many live conductors will go to feed each of the circuits

Here's a start

4 * 3 triple pole 3 phase motors will have 12 conductors
2 * 4 pole 3 phase motors will have 8 conductors
4 * number of live conductors in the pipe or trunking will be?
2 * number of live conductors in the pipe or trunking will be?

Total number of conductors entering the DB will be my invoice in pounds! :)
 
Last edited:
- 4 TP motors 4 x 3 =12
- 2 x TP&N motors 2 x 4 = 8
- 4 x single phase ring circuits 4 x 2 x 2 = 16
- 2 x single phase lighting circuits 2 x 2 =4

total 40.

how many live conducto
 

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